To solve the radical equation [tex]\( a = \sqrt{5a + 6} \)[/tex], let's proceed step-by-step.
1. Isolate the square root term:
[tex]\[
a = \sqrt{5a + 6}
\][/tex]
2. Square both sides of the equation to eliminate the square root:
[tex]\[
a^2 = ( \sqrt{5a + 6} )^2
\][/tex]
[tex]\[
a^2 = 5a + 6
\][/tex]
3. Rearrange the equation to standard quadratic form:
[tex]\[
a^2 - 5a - 6 = 0
\][/tex]
4. Solve the quadratic equation using the quadratic formula, [tex]\( a = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[
a = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(1)(-6)}}}}{2(1)}
\][/tex]
[tex]\[
a = \frac{{5 \pm \sqrt{{25 + 24}}}}{2}
\][/tex]
[tex]\[
a = \frac{{5 \pm \sqrt{49}}}{2}
\][/tex]
[tex]\[
a = \frac{{5 \pm 7}}{2}
\][/tex]
5. Find the two potential solutions:
For the positive case:
[tex]\[
a = \frac{5 + 7}{2} = \frac{12}{2} = 6
\][/tex]
For the negative case:
[tex]\[
a = \frac{5 - 7}{2} = \frac{-2}{2} = -1
\][/tex]
6. Check each solution in the original equation to verify it's valid:
For [tex]\( a = 6 \)[/tex]:
[tex]\[
6 = \sqrt{5(6) + 6}
\][/tex]
[tex]\[
6 = \sqrt{30 + 6}
\][/tex]
[tex]\[
6 = \sqrt{36}
\][/tex]
[tex]\[
6 = 6
\][/tex]
This solution is valid.
For [tex]\( a = -1 \)[/tex]:
[tex]\[
-1 = \sqrt{5(-1) + 6}
\][/tex]
[tex]\[
-1 = \sqrt{-5 + 6}
\][/tex]
[tex]\[
-1 = \sqrt{1}
\][/tex]
[tex]\[
-1 = 1
\][/tex]
This solution is invalid due to the square root producing a positive result, while the left side is negative.
Therefore, the real solution to the equation [tex]\( a = \sqrt{5a + 6} \)[/tex] is:
[tex]\[
a = 6
\][/tex]
